Two vertices of a triangle are the points $A(3,-1)$ and $B(5,7)$. Point $N(4,-1)$ is the intersection of the altitudes. Find equations for the sides of this triangle.
What I have done:
Let $h_1$ be the altitude from the vertex $A$, $h_2$ from the vertex $B$ and $h_3$ from the vertex $C$.
By definition, we have that:
$h_1 \perp \ BC, h_2 \perp \ AC, h_3 \perp \ AB$
so:
$$h_1=-\dfrac{1}{m(BC)}\cdot(x-x_A)+y_A=-\dfrac{x_C-5}{y_C-7}\cdot(x-3)-1$$
$$h_2=-\dfrac{1}{m(AC)}\cdot(x-x_B)+y_B=-\dfrac{x_C-3}{y_C+1}\cdot(x-5)+7$$
$$h_2=-\dfrac{1}{m(AB)}\cdot(x-x_C)+y_C=-\dfrac{1}{4}\cdot(x-x_C)+y_C$$
But when I use the orthocenter $N(4,-1)$ to find $x_C, y_C$ I dont get anything.
what should I do?
Best Answer
To solve the problem as stated, it's not necessary to write equations for the altitudes, or even to find point $C$. It just asks for equations representing the three sides of the triangle.
Side $AB$ is the simplest case, since points $A$ and $B$ are given. You have already computed its slope $m_{AB}=4$, and can use either $A$ or $B$ as the point on the line.
For side $AC$, notice that it must be perpendicular to the line $BN$. We don't need to know where these lines intersect to conclude that $m_{AC}m_{BN} = -1$. And $m_{BN}$ is simple to compute from the two given points $B$ and $N$. Knowing the slope and a point $A$ on the line, you can write the equation for $AC$.
Similarly, side $BC$ is perpendicular to the line $AN$, so you can compute $m_{AN}$ and from that determine $m_{BC}$, then write the equation for $BC$.